Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:16777216")
using namespace std;
using matrix = vector<vector<long long>>;
void FAST_IO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
}
const int MAXN = 2e5 + 100;
const double eps = 1e-7;
pair<int, int> seg[MAXN];
int pref[MAXN];
signed main() {
  FAST_IO();
  int sy, a, b;
  cin >> sy >> a >> b;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> seg[i].first >> seg[i].second;
    pref[i] = seg[i].second - seg[i].first;
    if (i) {
      pref[i] += pref[i - 1];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    double ans = 0;
    double k = (y - sy) / 1. / y;
    int l = 0, r = n, mid;
    while (l + 1 < r) {
      mid = (l + r) >> 1;
      int dx = seg[mid].first - x;
      double nx = x + dx * k;
      if (nx < a + eps) {
        l = mid;
      } else {
        r = mid;
      }
    }
    int dlx = seg[l].first - x;
    int drx = seg[l].second - x;
    double nlx = x + dlx * k;
    double nrx = x + drx * k;
    if (nlx < a + eps && a < nrx + eps) {
      if (b < nrx + eps) {
        ans += b - a;
        cout << ans << '\n';
        continue;
      }
      ans += nrx - a;
    }
    int L = -1;
    if (nlx < a + eps) {
      L = l + 1;
    } else {
      L = l;
    }
    l = 0, r = n;
    while (l + 1 < r) {
      mid = (l + r) >> 1;
      int dx = seg[mid].first - x;
      double nx = x + dx * k;
      if (nx < b + eps) {
        l = mid;
      } else {
        r = mid;
      }
    }
    dlx = seg[l].first - x;
    drx = seg[l].second - x;
    nlx = x + dlx * k;
    nrx = x + drx * k;
    if (nlx < b + eps && b < nrx + eps) {
      ans += b - nlx;
    }
    int R = -1;
    if (nlx < b + eps && b < nrx + eps) {
      R = l - 1;
    } else if (nrx < b + eps) {
      R = l;
    }
    if (R - L + 1 > 0) {
      int len = pref[R];
      if (L) {
        len -= pref[L - 1];
      }
      ans += len * k;
    }
    cout << ans << '\n';
  }
}